Question: Is ${651006}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {651006}= &&{6}\cdot100000+ \\&&{5}\cdot10000+ \\&&{1}\cdot1000+ \\&&{0}\cdot100+ \\&&{0}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {651006}= &&{6}(99999+1)+ \\&&{5}(9999+1)+ \\&&{1}(999+1)+ \\&&{0}(99+1)+ \\&&{0}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {651006}= &&\gray{6\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {6}+{5}+{1}+{0}+{0}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${651006}$ is divisible by $3$ if ${ 6}+{5}+{1}+{0}+{0}+{6}$ is divisible by $3$ Add the digits of ${651006}$ $ {6}+{5}+{1}+{0}+{0}+{6} = {18} $ If ${18}$ is divisible by $3$ , then ${651006}$ must also be divisible by $3$ ${18}$ is divisible by $3$, therefore ${651006}$ must also be divisible by $3$.